Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x-3y &= -3 \\ -3x-8y &= -5\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $-3x = 8y-5$ Divide both sides by $-3$ to isolate $x$ $x = {-\dfrac{8}{3}y + \dfrac{5}{3}}$ Substitute this expression for $x$ in the first equation. $3({-\dfrac{8}{3}y + \dfrac{5}{3}}) - 3y = -3$ $-8y + 5 - 3y = -3$ Simplify by combining terms, then solve for $y$ $-11y + 5 = -3$ $-11y = -8$ $y = \dfrac{8}{11}$ Substitute $\dfrac{8}{11}$ for $y$ in the top equation. $3x-3( \dfrac{8}{11}) = -3$ $3x-\dfrac{24}{11} = -3$ $3x = -\dfrac{9}{11}$ $x = -\dfrac{3}{11}$ The solution is $\enspace x = -\dfrac{3}{11}, \enspace y = \dfrac{8}{11}$.